Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury T = 435.5 \times 10^{-3} \frac{N}{m}.

Answers (1)

Surface tension energy, E=\sigma \Delta A

As we can infer from the law of conservation of mass, the volume of a drop V = v1 + v2

r_1 = 0.1 cm = 0.001 m

r_2 = 0.2 cm = 0.002 m

\Delta A = 4 \pi r_1^{2} + 4 \pi r_2^{2} - 4\pi R^{2}

\Delta A = 4 \pi \left [r_1^{2} + r_2^{2}- R^{2} \right ]

The new bigger drop has a radius of R,

\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi \left [ r_1^{3}+r_2^{3} \right ]

R^{3}= r_1^{3}+r_2^{3}

R^{3}= 9 \times 10^{-9}

R^{3}= 2.1 \times 10^{-3}m

Now, E=\sigma \Delta A

E = 4 \times 3.14 [(10^{-3})^{2} + (2 \times 10^{-3})^{2} - (2.1 \times 10^{-3})^{2}] \times 435.5 \times 10^{-3}

E = 1742 \times 3.14 \times 10^{-9} [0.59] = 5469.88 \times 0.59 \times 10^{-9}

E = 32.27 \times 10^{-7} J

Finally, the area will be smaller, when energy is released as the bigger drop is formed from smaller drops.

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